package com.sxkiler.demo.hard;

import org.junit.jupiter.api.Assertions;
import org.junit.jupiter.api.Test;
import java.util.*;
import com.sxkiler.demo.model.*;

/**
last-substring-in-lexicographical-order=按字典序排在最后的子串
<p>给你一个字符串&nbsp;<code>s</code>，找出它的所有子串并按字典序排列，返回排在最后的那个子串。</p>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>

<pre><strong>输入：</strong>"abab"
<strong>输出：</strong>"bab"
<strong>解释：</strong>我们可以找出 7 个子串 ["a", "ab", "aba", "abab", "b", "ba", "bab"]。按字典序排在最后的子串是 "bab"。
</pre>

<p><strong>示例&nbsp;2：</strong></p>

<pre><strong>输入：</strong>"leetcode"
<strong>输出：</strong>"tcode"
</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ol>
	<li><code>1 &lt;= s.length &lt;= 4 * 10^5</code></li>
	<li>s 仅含有小写英文字符。</li>
</ol>

 */
public class lastSubstring {
    

    class Solution {
        public String lastSubstring(String param0) {
            return null;
        }
    }

    @Test
    public void test(){
        Solution solution = new Solution();
        /**
        "abab"

        */
        //int [] num1 = new int[]{1,3};
        //int [] num2 = new int[]{2};
        //Assertions.assertEquals(solution.{{questionName}}(num1,num2),2);
    }
}

